Tìm \(x\ne0\)sao cho\(x\sqrt{x}=x+\sqrt{x}\)
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NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 7 2021
\(A=\left(\dfrac{\sqrt{x}+1+x+\sqrt{x}}{\sqrt{x}+1}\right)\left(\dfrac{\sqrt{x}-1-x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{-\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right)\left(\dfrac{-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\right)\)
\(=-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)=1-x\)
\(A=-1\Leftrightarrow1-x=-1\Rightarrow x=2\)
11 tháng 7 2021
a) Ta có: \(A=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=1-x
b) Để A=-1 thì 1-x=-1
hay x=2
23 tháng 10 2021
a) \(C=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{4\sqrt{x}}{x-1}=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(C=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}+1=3\sqrt{x}-3\Leftrightarrow2\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
12 tháng 9 2018
\(B=\frac{2+\sqrt{x}}{x-4\sqrt{x}+4}:\left(\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}+\frac{6-x}{x+2\sqrt{x}}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}+\frac{6-x}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)+\left(6-x\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{x\sqrt{x}-8+x+2\sqrt{x}+6\sqrt{x}-12-x\sqrt{x}+2x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{3x+8\sqrt{x}-20}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{\sqrt{x}\left(2+\sqrt{x}\right)^2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2\left(3x+8\sqrt{x}-20\right)}\)
\(B=\frac{\sqrt{x}\left(2+\sqrt{x}\right)^2}{\left(\sqrt{x}-2\right)\left(3x+8\sqrt{x}-20\right)}\)
tới đây mình bí rồi cậu làm giúp mình đi
mại dzo
NL
1
26 tháng 12 2021
\(\Leftrightarrow Mx^2=x^2-2x+\sqrt{2015}\\ \Leftrightarrow x^2\left(M-1\right)+2x-\sqrt{2015}=0\)
Ta có \(\Delta'\ge0\Leftrightarrow1+\sqrt{2015}\left(M-1\right)\ge0\)
\(\Leftrightarrow1+\sqrt{2015}M-\sqrt{2015}\ge0\\ \Leftrightarrow M\ge\dfrac{\sqrt{2015}-1}{\sqrt{2015}}\)
Vậy \(M_{min}=\dfrac{\sqrt{2015}-1}{\sqrt{2015}}\Leftrightarrow x=-\dfrac{b'}{a}=-\dfrac{1}{M-1}=\dfrac{-\sqrt{2015}}{\sqrt{2015}-1}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1 2021
\(A=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)
\(A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
\(y=4x^2+\dfrac{9}{x^2}-3\ge2\sqrt{\dfrac{36x^2}{x^2}}-3=9\)
\(y_{min}=9\) khi \(x^2=\dfrac{3}{2}\)
\(P=\dfrac{x-1}{4}+\dfrac{1}{x-1}+\dfrac{1}{4}\ge2\sqrt{\dfrac{x-1}{4\left(x-1\right)}}+\dfrac{1}{4}=\dfrac{5}{4}\)
\(P_{min}=\dfrac{5}{4}\) khi \(x=\dfrac{3}{2}\)
CW
22 tháng 12 2017
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}}{\sqrt{x}-1}\)
\(x\sqrt{x}=x+\sqrt{x}\)
\(\Rightarrow\sqrt{x}\left(x-\sqrt{x}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\x-\sqrt{x}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\left(l\right)\\x-\sqrt{x}-1=0\left(1\right)\end{cases}}}\)
Từ (1) . Có: \(\Delta=\left(-1\right)^2-4\left(-1\right)=5>0\Rightarrow\sqrt{\Delta}=\sqrt{5}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\left(n\right)\\x=\frac{1-\sqrt{5}}{2}\left(n\right)\end{cases}}\)
Vậy \(x=\left\{\frac{1+\sqrt{5}}{2};\frac{1-\sqrt{5}}{2}\right\}\)
đặt \(\sqrt{x}\)\(=y\)
=> \(x=y^2\)
ta có pt : \(y^3=\)\(y^2+y\)
\(\Leftrightarrow y^3-y^2-y=0\)
\(y\ne0\)\(\Rightarrow\)\(y^2-y-1=0\)
ptdttnt rồi bạn tự tính y => x